A cornerstone of physics may require a rethink if findings at the National Institute of Standards and Technology (NIST) are confirmed. Recent experiments suggest that the most rigorous predictions based on the fundamental theory of electromagnetismone of the four fundamental forces in the universe, and harnessed in all electronic devicesmay not accurately account for the behavior of atoms in exotic, highly charged states.

The theory in question is known as quantum electrodynamics, or QED, which physicists have held in high regard for decades because of its excellent track record describing electromagnetism's effects on matter. In particular, QED has been especially useful in explaining the behavior of electrons, which orbit every atomic nucleus. But for all of QED's successes, there are reasons to believe that QED may not provide a complete picture of reality, so scientists have looked for opportunities to test it to ever-increasing precision.

One way to test parts of QED is to take a fairly heavy atomtitanium or iron, for exampleand strip away most of the electrons that circle its nucleus. "If 20 of titanium's 22 electrons are removed, it becomes a highly charged ion that looks in many ways like a helium atom that has been shrunk to a tenth its original size," says NIST physicist John Gillaspy, a member of the research team. "Ironically, in this unusual state, the effects of QED are magnified, so we can explore them in more detail."

Among the many things QED is good for is predicting what will happen when an electron orbiting the nucleus collides with a passing particle. The excited electron gets bumped up momentarily to a higher energy state but quickly falls back to its original orbit. In the process, it gives off a photon of light, and QED tells what color (wavelength) that photon will have...

Interesting article, but it's confusing. If you take an atom of iron or titanium and strip away most of its electrons, have you not fundamentally changed the structure of the atom sufficiently that it has now become something else - like helium - as the article suggests? If that's the case, why not use an atom of helium instead of converting another atom into something that looks or acts like helium under these conditions??

Hopefully, someone with more chemistry or physics background can point me to an explanation.

I’m not an electrician but it’s always seemed to me
to be somewhat anachronistic to use atomic power to
run a steam turbine to make electricity. Surely there
must be a way of direct production...? Here we are
generating electrical current with the same means
of production used from the very beginning.

8
posted on 12/03/2012 3:00:33 PM PST
by tet68
( " We would not die in that man's company, that fears his fellowship to die with us...." Henry V.)

Hard to come up with a more efficient way of converting heat to electricity at that scale and reliability. If you’re trying to capture electrons directly and send them out the transmission grid you’re going to be importing truckloads of nuclear fuel every day, not to mention running the biggest X-Ray machine in the world.

“If 20 of titanium’s 22 electrons are removed, it becomes a highly charged ion that looks in many ways like a helium atom that has been shrunk to a tenth its original size,” says NIST physicist John Gillaspy, a member of the research team.
***And if that heavy metal is absorbing hydrogen atoms, the resultant Condensed Matter starts behaving as if it were under a plasma, with far more likelihood of generating fusion events than previously.

The energy levels that the electrons can possess are a function of their attraction to the nucleus, which is largely a function of the nuclear charge (which equals the number of protons).

As long as you don’t change the number of protons, you don’t change the identity of the element. An iron nucleus identifies it as iron whether it has all it’s electrons (as in metallic iron) or whether it’s missing three of it’s electrons (as it the form of iron found in iron oxide / rust).

Losing 20 out of 22 electrons is a pretty big perturbation, but as long as the number of protons in the nucleus stays the same, you’re dealing with the same element.

The experimental results they came up, if correct, suggest there may be some electron-electron repulsion, or orbital-orbital perturbation they hadn’t accounted for in determining orbital energies via QED theory. But that’s way beyond any physics I can make sense of.

SuperSymmetry and QED aren't comparable in any sense. SS has always been speculative; it's attractive for some amthematical reasons but not mainstream. QED is as mainstream in particle physics as you get. It's not too far off to call it "The Central Dogma" (as molecular biologists refer to DNA.)

But there's nothing in this article that indicates that QED is in trouble, only that photomeission wavelengths differ from those predicted purely by QED. A much more likely explanation than the BIG HEADLINE that QED "might" be wrong is that there are other energy effects not accounted for by the experimenters.

That doesn't grab headlines or generate grant money, so they aren't going to broadcast the more pedestrian (and likely) explanation.

The energy levels that the electrons can possess are a function of their attraction to the nucleus, which is largely a function of the nuclear charge (which equals the number of protons).

Only true in hydrogenic atoms. The energy levels of multiple electron atoms are dominated by the Exclusion Principle.

As long as you dont change the number of protons, you dont change the [nuclear] identity of the element

True in terms of the physics, but not the chemistry. An He+ ion does not behave like Helium for chemical purposes, (though Z is the same) and it certainly doesn't behave like Hydrogen (though Ne is the same.)

An iron nucleus [nucleus, yes!] identifies it as iron whether it has all its electrons (as in metallic iron) or whether its missing three of its electrons (as it the form of iron found in iron oxide / rust).

Again, too oversimplified to be true. There is significant overlap of the state vectors of the oxygen and iron in rust, and the delocalized electrons are in molecular orbitals; they are not ions, and they are not atomic orbitals, either. If oxidized iron identified as iron when three of its electrons are significantly delocalized into molecular orbitals, there would be no reason for rust to form (and it would not form.) It is iron in terms of the nuclear chemistry, but is no longer iron in terms of atomic (ordinary) chemistry.

Losing 20 out of 22 electrons is a pretty big perturbation

It's not a perturbation from that perspective, and a perturbation approach wouldn't be done that way: the perturbation would treat the atom as a hydrogenic atom with Z=22, and one electron. The second electron is then added as a perturbation. Just one additional electron is still a very large effect.

I know you will think I am picking nits; however, in terms of what's reported here, the researchers are essentially examining the atomic (not nuclear) properties of certain atoms. For the purposes of these studies, what they're saying is: we expect the emission spectra to look like an "overcharged" hydrogen or helium nucleus. The energy levels of those are very well known (the former -- hydrogen -- is, in fact, exactly calculable, with any Z you want, and is an exercize for junior level nuclear physics or P-chem courses at undergraduate level. With a hydrogenic wavefunction with Z=22, the Schroedinger equation doesn't apply very well. Still, the values with a full relativistic treatment with QED are calculable to a very high degree of precision.)

So ... quite surprising. I bet it turns out to be an experimental artifact and is not new science.

No problem with the “nits” - they were well-picked. At what level one presents this stuff depends a lot on what audience you’re shooting at. Given the nature of the original question, I’m not sure delocalized MO’s and the wave equation would be the best place to start - but I do appreciate your well-presented elaborations.

"Ere many generations pass, our machinery will be driven by a power obtainable at any point in the universe. This idea is not novel... We find it in the delightful myth of Antheus, who derives power from the earth; we find it among the subtle speculations of one of your splendid mathematicians...Throughout space there is energy. Is this energy static or kinetic? If static our hopes are in vain; if kinetic -- and this we know it is, for certain -- then it is a mere question of time when men will succeed in attaching their machinery to the very wheelwork of nature." -- Nikola Tesla, American Institute of Electrical Engineers address, 1891

The electrons are those grey things orbiting around; the red and blue balls in the center are the neutrons and protons.

Electrons have a -1 charge; each proton a +1 charge; and the neutrons are neutral (hence the name).

In any element -- that is, when left to itself, you haven't reacted it or ionized it or anything -- the atom has a 0 charge, since the number of protons and electrons is identical.

Isotopes (like the deuterium in heavy water, or like Carbon-14) differ from the ordinary element because they weigh more: but the extra weight comes from extra neutrons in the nucleus, so they are still neutral.

The trick is, when you pull away so many electrons from iron or titanium, the charge of the nucleus is still the same: so you have a whole bunch of positive charge in the middle, pulling in on just a couple of electrons: since opposites attract, the electrons move in closer to the nucleus, so the atom is "smaller" -- kind of like as the Solar System would be smaller if you got rid of every planet past Mars. (Conceptually, not mathematically, that is, since the planets are attracted by gravity, not charges...)

Full Disclosure: I have a PhD in molecular physics.

Cheers!

24
posted on 12/03/2012 6:59:00 PM PST
by grey_whiskers
(The opinions are solely those of the author and are subject to change without notice.)

Does QED allow for relativistic effects? I recall hearing at a theoretical chemistry conference, that the color of gold was due to relativistic effects on the inner electrons; presumably, using a similarly heavy highly ionized species to "mimic" helium by drawing in the electrons, would result in very high orbital velocity; hence, room for relativistic effects?

Cheers!

25
posted on 12/03/2012 7:02:10 PM PST
by grey_whiskers
(The opinions are solely those of the author and are subject to change without notice.)

It's not a perturbation from that perspective, and a perturbation approach wouldn't be done that way: the perturbation would treat the atom as a hydrogenic atom with Z=22, and one electron. The second electron is then added as a perturbation. Just one additional electron is still a very large effect.

I thought perturbation theory for electronic structure was supplanted by configuration interaction; and that people generally liked to use Density Functional Theory rather than actually solving for the wavefunction explicitly.

Or am I just having a senior moment?

Cheers!

26
posted on 12/03/2012 7:06:27 PM PST
by grey_whiskers
(The opinions are solely those of the author and are subject to change without notice.)

I was taking his observation at face value: if you regard this as a perturbation how would you approach it.

However, for hydrogenic state vectors DFT is not needed; as you know, there's an exact solution, and for helium a perturbation is OK. I haven't taught it for a long time, but IIRC, the trick is usually tuning a Zeff for each principal quantum number (n) then the two particle wave function "isn't too bad" even for angular momentum quantum numbers that have significant penetration into the inner shells.

Of course, that starts to crap out very quickly -- like, at Lithium -- so there isn't much of even pedagogic value for multiple electron systems in perturbation theory.

QED is fully relativistic. Historically it derived from second quantization of the Dirac equation. Recall that anti-matter came out of the Dirac equation for the electron. So did spin. In the Schroedinger/Heisenberg version of quantum mechanics neither of those things really exist. Spin was just thrown in ad hoc because it was known to be there. But you can't produce it from first principles without relativistic treatment. Anti-matter kinda surprised even Dirac. But he stuck to his guns. You can't really do pair-production, negative energy, or virtual particles in QED without a relativistic basis.

Nobody likes to remember "second quantization" anymore; it's like the Klingons in the original Star Trek. Kind of embarrassing to High Energy guys these days. I was never a High Energy Physicist. I just found it interesting. Naturally my adviser regarded courses beyond the candidacy exam not related to my field as a waste of time.

Gold. Now you are really testing my memory. But ... I do not think so. I seem to recall that if you actually do a wild-ass guess (that works surprisingly well, kind of like that neutron/proton stability model that comes from viewing the nucleus as a particle-in-a-box) you find that the inner electrons in all elements from the Lanthanides and beyond have enough energy that they have to be treated relativistically.

Gawd that is a long time ago... anyway, no, color in pure metals is a result of band theory. I believe for most metals, there are so many energy levels so close at the top of the conduction band that they radiate essentially continuously (thus appear silver, as most metals do.) Copper and Gold have some weirdness in the Fermi-levels near the top of their conduction bands that cut-off blue light.

I'll see if I can dig up a web page in a while; must help daughter with math.

Thank you. While I grasp the concept of the nucleus remaining the same without all of those electrons that have been pulled out (I squeaked through high school chemistry, but a few lessons stuck! LOL!), what ultimately changes in the atomic structure as the result of those missing electrons? Is it less able to bond with other atoms in a molecular structure, or did I sleep through that part of ny high school chemistry class?

Its bonding properties change. In this particular case with so many electrons gone it is going to become strongly electronegative. Like a super-duper Oxygen or Florine. It will be pulling electrons away from everything within reach.

Oh geez. You are 100% correct! I had forgotten that the Dirac equation leads to a completely different aufbau than the Schrodinger/Heisenberg one. In addition to the Lanthanide contraction caused by relativistic effects (at least I remembered that part correctly) in Au there is an additional contraction of the 6s orbital so the 5d -> 6s transition (into the conduction band) is so small it absorbs in the blue, rather than UV and higher as is the case of the bound electron orbitals in most metals. Have a go: http://voh.chem.ucla.edu/vohtar/fall02/classes/172/pdf/172rpint.pdf.

A more lay accessible treatment is here: http://www.fourmilab.ch/documents/golden_glow/. Which has some errors (the transition is 4d->5s in Ag, not Au) but omits the complexity of band theory altogether; and seems "correct enough."

Now, that still doesn't explain Cu, and I sure don't remember any more. No time tonight, though.

Ok, I’m staying up with you so far (and removing a LOT of dust that settled on the high school chemistry information!), but I keep getting hung up on the atomic weight. Is the atomic weight based solely on the nucleus? Do the atoms have zero weight or is it too negligible to make any difference?

BTW, thanks for your patience in helping me understand this stuff. I really enjoy learning new things, no matter how many times I have learned it before!!

The neutron and proton masses are both close to 940 MeV/c^2. [Neutron is actually slightly heavier.] The electron is 0.512 MeV/c^2. So, it takes ~2000 electron masses to make a nucleon mass. Since the most massive element has only slightly more than 100 electrons, the electron mass is always negligible compared to nuclear mass (in chemistry.) The biggest variation in nuclear masses actually come from averaging isotope species and (to a much smaller extent) differences in binding energy.

“Im not an electrician but its always seemed to me
to be somewhat anachronistic to use atomic power to
run a steam turbine to make electricity.”

Bingo. The giant energy problem of the next decade(s) ISN’T where are we going to find stuff to burn.

The problem is: Where, on a planet where only one percent of the water on it is fresh water, are we going to find sufficient water for both energy exploration AS WELL AS production.

Consider this, however:

A generator is nothing more than a piece of iron spinning inside a magnetic field. The core of the earth is made up of nickel and iron. It rotates within a giant magnetosphere. Where does that electrical current manifest itself? Does that current require water to produce?

Correction: The biggest variation in nuclear masses atomic weights actually come from averaging isotope species and (to a much smaller extent) differences in binding energy. [Nuclear mass itself only changes as a result of binding energy.]

QED is fully relativistic. Historically it derived from second quantization of the Dirac equation.

Feh. That's what I get for relying on Feynmans QED: The Strange Theory of Light and Matter (ISBN-10: 0691024170 | ISBN-13: 978-0691024172, $7.77 at Amazon.com) instead of having taken a real course in it.

As far as the Gold, I'll take your word for it: it was an offhand comment during a discussion section after a paper at a long-gone Theoretical Chemistry conference, but it was odd enough that it lodged in the esophagus of my mind.

Incidentally, speaking of the Dirac equation, doesn't it leave room for magnetic monopoles? John Van Vleck would be deeply saddened...

Cheers!

42
posted on 12/04/2012 4:06:28 PM PST
by grey_whiskers
(The opinions are solely those of the author and are subject to change without notice.)

Incidentally, speaking of the Dirac equation, doesn't it leave room for magnetic monopoles?

Well ... it could, but he never developed a fully formed Hamiltonian that would include magnetic monopoles in his theory of the electron, AFAIK.

The famous Dirac/Magnetic Monopoles thing you're talking about was that he published a symmetrized revision of Maxwell's Equations which included magnetic monopoles complete with a magnetic Gauss's Law and Magnetic displacement currents. The whole deal. The consequences were interesting. But it was purely classical.

As you correctly point out, all that is required is a conductor moving in an electromagnetic field (specific conductors, such as iron and nickel are not required.) Or [what relativistically amounts to the same thing] a stationary conductor inside a time varying electromagnetic field will also function as a generator.

The expansion of water from the nuclear pile's heat is just a way to move the turbine. That's all. Any way you could convert the heat into some way of moving the conductor through the field works just fine. As it turns out, for many reasons including the enormous amounts available and its low toxicity, water is actually a pretty good fluid to use.

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